QA 

371 

M4 


DIFFERENTIAL 
EQUATIONS 


iNJEERING   & 

jraryI 


ol 

01 
0; 

1    : 

6 
9 

5 
3 
3 


MAURUS 


^^^^^^^H 

^^^^Hii7iT7;.j^r7^,i:  •-.;:;  :,..,>.  •.-« 

^^^^^^H 

^HHft'|'l?-::$M|li 

HIH| 

^^piiiii 

[ifornia      H 

^^^B';J||;:k.:'v|| 

onal           H 

^^^BA:liiBI§^c-' 

lity             1 

^H|--:-:;:|:H;;v 

k  is  DUE  on  U; 


m 


Digitized  by  tine  Internet  Arcliive 

in  2007  witli  funding  from 

IVIicrosoft  Corporation 


littp://www.arcliive.org/details/elementarycourseOOmauriala 


AN  ELEMENTARY  COURSE  IN 
DIFFERENTIAL  EQUATIONS 


BY 

EDWARD  J.  MAURUS,  M.S. 

PBOFESSOR  OF  MATHEMATICS  IN  NOTRE  DAMB  UNIVEB8ITT 


GINN  AND  COMPANY 

V  YORK    •     CHICA 
IS     ■     COLUMBUS 

3937^ 


BOSTON     •     NEW  YORK    •     CHICAGO    •    LONDON 
ATLANTA    •    DALLAS     ■     COLUMBUS    •    SAN  FRANCISCO 


COPYRIGHT,  1917,  BT 
EDWARD  J.  MAURU8 


AI.I.  BIOHT8  BESEBVBD 
117.10 


gbe  gtbengnm  >re« 

GINN  AND  COMPANY  •  PRO- 
PRIETORS •  BOSTON  •  U.S.A. 


Enjfrccring  & 
Mb  J;:matic8l 

Scionces 

tibraiy 


PREFACE 

The  aim  of  the  author,  in  preparing  this  work,  has  been 
to  afford  his  classes  an  easy,  condensed  course  in  ordinary 
differential  equations,  and  to  serve  as  a  review  of  Integral 
Calculus. 

With  few  exceptions,  the  numerous  problems  are  new, 
though  fashioned  after  the  old  models. 

EDWARD  J.  MAURUS 
NoTBE  Dame,  Ikdiana 


iii 


CONTENTS 

CHAPTER  PAGE 

I.  Definitions.   Derivation  of  a  Differential  Equa- 
tion   1 

11.  Differential  Equations  of  the  First  Order  and 

First  Degree 4 

III.  Differential  Equations  of  the  First  Order  but 

NOT  of  the  First  Degree 14 

IV.  Differential  Equations  of  Orders  Higher  than 

the  First * 19 

ANSWERS 43 


AN  ELEMENTARY  COURSE  IN 
DIFFERENTIAL   EQUATIONS 


vii 


DIFFERENTIAL  EQUATIONS 


CHAPTER  I 

DEFINITIONS.     DERIVATION  OF  A  DIFFERENTIAL 
EQUATION 

Any  equation  containing  differentials  or  derivatives 
is  called  a  differential  equation.  If  only  one  independ- 
ent variable  is  involved,  the  equation  is  an  ordinary 
differential  equation.  The  following  are  examples  of 
ordinary  differential  equations: 

X—  +  y  =  xy  cot x.  (1) 

dx 

e^dy  +  ye^dx  —  dy.  (2) 

h(S)T=-(S)- 

The  order  of  a  differential  equation  is  the  order  of 
the  highest  derivative  in  the  equation,  —  for  example, 
(1),  (2),  and  (3)  are  of  the  first  order;  (4)  and  (5) 
are  of  the  second  order. 

The  degree  of  a  differential  equation  is  the  degree  of 
the  derivative  of  highest  order  in  the  equation,  —  for 

1 


2  DIFFERENTIAL  EQUATIONS 

example,  (1),  (2),  and  (4)  are  of  the  first  degree ;  (3) 
and  (5)  are  of  the  second  degree. 

A  solution  of  a  differential  equation  is  an  equation  of 
relationship  between  the  variables,  free  from  differen- 
tials and  derivatives,  which  will  satisfy  the  differential 
equation.  In  the  process  of  solving,  one  or  more  inte- 
grations must  be  performed,  each  involving  an  arbitrary 
constant.  The  general^  or  complete,  solution  thus  contains 
an  arbitrary  constant  for  each  unit  in  the  order  of  the 
equation ;  that  is,  the  complete  solution  of  a  differential 
equation  of  the  second  order  will  contain  two  arbitrary 
constants,  etc.  The  differential  equation  may  be  con- 
sidered as  derived  from  its  complete  solution  by  the 
elimination  of  the  arbitrary  constants. 

Example  1.  Form  a  differential  equation  by  eliminating  the 
constant  c  from  the  equation  xy  =  c  sin  x.  Since  two  equations 
are  required  to  eliminate  one  arbitrary  constant,  a  second  equa- 
tion is  needed.  This  may  be  formed  by  differentiating  the  given 
equation :  xdy  +  ydx  =  c  cos  xdx.   Eliminating  the  c,  we  have  the 

differential  equation  x  ~  +  y  =  xy  cot  x. 
ax 

Example  2.  Form  a  differential  equation  by  eliminating  the 
constants  Cj  and  Cg  from  the  equation  y  =  c^e^  +  c^e^^.  Since 
there  are  two  constants  to  eliminate,  three  equations  are  needed. 
The  two  extra  equations  are  formed  by  differentiating  the  given 
equation  twice,  giving 

dy  d^v 

-2-  =  c.e"  +  2  CaS^^     and     — ^  —  c.e^  +  4  c^e^^. 

dx  dx^ 

Eliminating,  by  determinants  or  otherwise,  we  have 


DEFINITIONS  3 

EXERCISE  1 

Eliminate  the  constants  from  the  following  equations : 

1.  a^  +  -if  =  ex.  6.  y  =  c^e'  +  c^e^^  +  c^e^". 

2.  mi/ =  m^x  +  4:.  7.  ev  =  c^e''  +  c^e-'^. 

3.  i/=3(:^x-\-  c^.  8.  y  =  Cj  sin  8  X  +  Cj  cos  8  x. 

4.  y  =  c^e^^  +  c^e^^.  9.  y  =  \  sec x  +  c^  tan x. 

5.  y  =  c^logx  +  c^.  10.  y=c^x-\/l—^+c^(l—27?). 

11.  (x-a)2+(2/-/8)^=25. 


CHAPTER  II 

DIFFERENTIAL  EQUATIONS  OF  THE  FIRST  ORDER  AND 
FIRST  DEGREE 

Case  I.  When  the  variables  can  be  separated,  the 
solution  becomes  a  problem  in  direct  integration. 

Example.    Solve  the  equation  x  dy  —  y  dx  =  dx  —  dy. 
Transposing  and  uniting,    (x  -]rl)dy  =  (y  +  1)  dx. 

Dividing  by  (X  +  1)  (^  +  1),       -^  =  -^. 

y  +  I      X  +  L 

The  variables  are  now  separated. 

Integrating,  log  (^  +  1)  =  log  (x  + 1)  +  c. 

The  (?  is  an  arbitrary  constant  of  integration  and  may 
be  put  in  various  forms  suitable  to  the  problem.  The 
simplicity  of  the  solution  generally  depends  upon  the 
selection  of  a  proper  form  for  the  c. 

In  the  example  given,  log  c  is  as  much  an  arbitrary 
constant  as  c.  Replacing  the  c  by  logc,  the  answer 
assumes  the  form  t 

log(3/+l)  =  log(a;+l)  + logc     or     y  +1=  c(x  +  l}. 

EXERCISE  2 
Solve: 

1.  2dx  ■{■  xdy  =  0.      3.  cos^ydx  +  cos^xdy  =  0. 

dy  4.  \iaiTi X  GOS^ y dx+tsinyaos^ xdy  =0. 

2«    ,    +  2  XM  =  0. 

dx  "  5.  xdy  +  ydx  =  0. 

4 


FIRST  ORDER  AND  FIRST  DEGREE  5 

6.  xydx  -\-  dy  =  x^dy. 

7.  tan?/  cos^ydx  +  tanx  cos^ xdy  =  0. 

8.  e'dy  +  ye'dx  =  dy. 

9.  sin  x  cos  ydx  -\-  cos  a?  sin  ^/c^y  =  0. 

10.  cos^xdy  +  cos^ydx  =  cos^a;  eos^ydx. 

11.  ic'^rf^/  -{•  y^dx  -\-  dx  -\-  dy  =  0. 

12.  iccZy  =  Vy^+ltte. 

13.  a;  v^  +  y  =  a;v  cot  cc. 

ace 

14.  y <ia;  —  cc^y  =  (ic^  + 1) c?y  —  2  xydx. 

15.  a^ydy  +  cc^/^c^a;  +  xt/a;  +  ?/c?y  =  0. 

16.  (x"" -Tx  +  12)^  +  xy  -  5y  -  2x  -^10  =  0. 


17.  (l-ycot2/)g  +  ^  =  0 


dx      X 

Case  II.  If  the  equation  is  homogeneous  in  x  and 
3/,  substitute  ^  =  vx.  Then  separate  the  variables  and 
solve  by  Case  I. 

Example.    2  x^dy  =  ^  x^ydz  —  y^dy. 

Put  y  =  vx;  then  dy  =  vdx  +  x dv. 
Substituting  in  the  given  equation, 

2  vx^dx  +  2  ar*dy  =  3  vxMx  —  v*x^dx  —  v^x*dv. 

Dividing  by  x%    2  vdx  +  2  xdv  =  3  vdx  —  v*dx  —  v^xdv. 

Separating  the  variables, 

dx  ,      v^  +  2     ^        „ 

f- z ~  dv  =  0, 

X       v(^v^  —  1) 

dx  .      dv         2dv  ,  (2  r  +  1)  dv      _ 
or  —  + +  — T-j- — ±j-  =  0. 

X  V  —  1  V  V^  +  V  +  I 


6  DIFFERENTIAL  EQUATIONS 

Integrating,  log  x  +  log  (j;  —  1)  —  2  log  v  +  log  (v^  +  r  + 1)  =  log  c. 
Uniting  and  dropping  the  logarithms. 

Substituting  -  =  v,         x^  —  y^  =  cy^. 
EXERCISE  3 

Solve : 

1.  (2x  +  3 y)cia;+ (4a;  +  6 2/)(^2/  =  0. 

2.  (4x  +  5y)cia;+(6a;+7y)c?y  =  0. 

3.  (2a;  — 3y)c?ar +  (4a;  — 5y)c?y  =  0. 

4.  (ar"  +  y^  c?a;  +  xyrfy  =  0. 

5.  (^  —  y^^dx  -\-  xydy  =  0. 

6.  {x^  +  2xy  +  Sf)dx-\-{x^-\-&xy  +  5'f)dy  =  Q. 

7.  x<^ -f- icc?y  +  2?/c?a;  =  0. 

8.  xdx  +  2/c?2^  =  (x^  +  y^)^dx. 

9.  2xy(ia;-(2a^  +  2^)<i2^  =  0. 

10.  (2ar*  +  6xy  +  3?/^c?a;+(3ar'  +  6a;y+7y^(£y  =  0. 

11.  a;»-2r*  +  2xy^  =  0. 

12.  {7*  —  2f)dx  +  Zxfdy  =  0. 

dy  _     3ar'y  dy  _y{2x^  +  f) 

•  eia;      3a;»  +  /'  *  rfa;      a;(a;«  +  2y«)' 

Case  III.    If  the  equation  is  of  the  form 

(a^x  +  \y  +  <?j)  rfa;  +  (agO;  +  igt/  +  c^  dy  =  0, 
it  is  called  nonlwmogeTieous  of  the  first  degree.    Substitute 
x  =  3^  -\-Xq  and  y  =  y'-\-  y^.    Then  dx  =  dx/  and  dy  =  c?y', 
and  the  equation  becomes 

{<h^  +  hy"  +  <h^Q  +  hVo  +  Ci)  <^a/ 

+  {a^  +  h^'  +  a^Q  +  h^Q  +  c^)  dy'  =  OK 


FIRST  ORDER  AND  FIRST  DEGREE  7 

Place  a^XQ  +  b^T/^  +  Cj  =  0  and  a^Q  +  b^Q -\-c^=0.  If  these 
equations  are  not  identical  or  contradictory,  they  can 
be  solved  for  Xq  and  i/q.  These  values,  being  substituted 
in  equation  (a),  reduce  it  to 

(a^x^  +  b^y'}  dx'  4-  (a^x'  +  b^i/'}  di/'  =  0, 

which  is  homogeneous,  and  solvable  by  Case  II. 

If  the  auxiliary  equations  are  either  identical  or  con- 
tradictory, place  a-^x  +  b-^i/  =  v,  eliminating  x  or  y,  and 
solve  the  resulting  equation. 


EXERCISE  4 

1.  (2x-^Sy  +  4:)dx-\-(4x  +  5y  +  6)dy  =  0. 

2.  (x  +  3i/  —  2)dx  -{-  (5x  -{-  7 y  —10)dy  =  0. 

3.  (x-2y-^-3)dx  +  (3x-6y-]-10)dy  =  0. 

4.  (6x  -  Sy  -\-l)dx  -\-(9x  -12y  -\-10)di/  =  0. 

5.  (5x-^6y-{-8)dx  +  {7x-{-Sy-\-10)dy  =  0. 

6.  (x  +  2y-\-3)dx+(4:X  +  5y  +  6)dy  =  0. 

7.  {3x-^y  —  5)dx+(5x-6y  —  7)dy  =  0. 

8.  (4:x  —  5y  +  6)dx—(5x  +  3y  —  ll)dy=0. 
dy_7x  —  9y  +  7 


dx      9x-{-2y  +  9 

dy  _     3x  —  5y  -\-  S 
dx~  21x-35y-\-72' 


10.  -^  = 


Case  IV.  A  differential  equation  formed  from  its 
primitive  by  differentiation  without  reduction  is  called 
an  exact  equation.  If  it  is  of  the  first  order,  it  can  be 
put  in  the  form  Mdz  +Ndy  =  0. 


8  DIFFERENTIAL  EQUATIONS 

Example  1.    xeV  +  2  x^y  +  3  y  cos  x  ■\-  y^  =  k. 

Differentiating,  we  obtain 

(e»  +  6  ar^y  -  3  y  sina:)rfx  +  (ze"  +  2 a:»  +  3  cosx  +  3  ^)rfy  =  0. 

Call  the  original  expression  u.    Then  il/  =  e"  +  6  x^y  —  S  y  sin  x, 

which   is  —  >  and  N  =  xe"  +  2 x^  +  3  cos x  +  Sy\  which  is  — . 
8x  dy 

Since  = 1  the  test  for  an  exact   differential  equation 

8ydx      dxdy  g^  ^^ 

of  the  first  order  is  as  follows :  Obtain  —  and If  these 

results  are  identical,  the  equation  is  exact.    " 

To  solve :  Integrate  Mdx  with  respect  to  x.  This  will 
give  all  the  terms  of  the  solution  which  contain  x.  If 
the  solution  has  any  terms  not  containing  x,  they  will 
not  appear  in  Mdx,  but  will  appear  in  Hdi/,  from  which 
they  may  be  obtained. 

Example  2.   (2  ax  +  2  hy)  dx  +  (2hx  +  2  by)  dy  =  0. 

Here  M=  2ax  +  2hy     and     N=2hx  +  2by. 

—  =  2h  and  =  2  A. 

dy  ox 

Hence  the  equation  is  exact. 

Integrating  the  first  part  with  respect  to  x,  we  have  ax^  +  2  hxy. 
This  gives  all  the  terms  involving  x.  The  second  part  contains 
one  term,  2  bydy,  without  an  x.  Integrated,  this  gives  by^.  Hence 
the  complete  solution  is  ax^  +  2  hxy  +  by^  =  c. 

EXERCISE  5 

1.  {4:X-3i/)dx-(3x-Sy)di/  =  0. 

2.  (7 x  -\-  Sy)dx  -\-(Sx  +  9y)dy  =  0. 

3.  (2x  +  3y)dx+(3x  +  Uy)dy  =  0. 

4.  (x^+2xy)dx  +  a^dij=0. 

5.  (2x''  +  6xij  +  3if)dx  -^(Sx'^  +  6x1/  +  7y^dy  =  0. 


FIRST  ORDER  AND  FIRST  DEGREE 

6.  xy^dx  -f-  ^ydy  -\-  xdx  +  ydy  =  0. 

7.  (4  a;8y  -  3  x'y'^ dx  +{x*-2  x^y) dy  =  0. 

8.  sin  a;  cos  ydx  +  cos  x  sinydy  =  0. 

9.  (Sx'y-Sey -2xe^v^ys^dx 

+  (a;«  -3xey-  SxV^'  +  3xf)dy  =  0. 

10.  tany  •  c?a;  +  a;  sec^ydy  =  0. 

1     ,  ^dy  _ 

11.  sm-^yd'x  +  =  0. 


12  # nydx^^ 

'  (x  +  iy     (x  +  iy'+'^ 

13.  e'^  (2  e^  +  e'J)  dx  +  e^'  (e^  +  2  e?')  c^y  =  0. 

14.  2  a;'^^/^/  4-  4  a;?/(ix  =  e^dy  —  e^dx  +  ye^dx  —  xe^dy. 

xdx  +  ydii 

15.  — ,       ^   -^  =  fZa;. 


16. 


■Vx'+tf 
2  xdx  — 2  ydy  _  xdy  -\-  ydx 


t  ^y 


Case  V.  If  the  differential  equation  has  been  re- 
duced after  differentiation,  it  may  no  longer  be  exact. 

Example.  2  x^  —  3  x^y^  =  c.  (1) 

Different]  ating, 

(6  xV  -  15  xSfydx  +  (8  x^'f-  6  ifiy)  dy  =  0.  (2) 

This  is  exact.    Dividing  by  x^y, 

(6  f  - 15  x'^y)  dx  +  (8  xy^  -  6  x^)  dy  =  0.  (3) 

Here  M=Qy^-15 x^y     and       iV  =  8 xy^  -  6  x^, 

and  —  =  18w2-15x2     and     —  =  Sy^-lSx^ 

dy  dx 

Hence  equation  (3)  is  Twt  exact.  The  factor  a?i/,  which 
would  reduce  equation  (3)  to  (2)  and  make  it  exact,  is 
called  an  integrating  factor.    Sometimes  the  integrating 


10  DIFFERENTIAL  EQUATIONS 

factor  for  a  given  equation  can  be  readily  found  by  in- 
spection.   Various  rules  might  be  given  for  other  cases. 
The  following  will  prove  useful  in  many  problems : 
Multiply  the  given  equation  by  x'^y^    In  the  new 

equation  get  -;—  and  -— -•    If  the  method  is  possible, 
oy  ex 

similar  terms  will  be  found.  Equate  the  corresponding 
coefficients  and  solve  for  m  and  n, 

EXERCISE  6 

1.  (9ar+14y)da;+7a;rfy  =  0. 

2.  {x  ■j-2y)dx  4-  xdy  =  0. 

3.  (12  xy  - 15  f)  dx  +  (10  x"  - 18  xy)  dy  =  0. 
o     4.  (16xy-15i/^dx-\-(Sx'-21xy)dy  =  0. 

'  dx       x^  —  2xy 

6.  (5xy  —  2 y^dx  +  (5x'^  —  4: xy)dy  =  0. 

7.  (35xy  +  20y'')dx-\-{lSx'  +  33xy)dy  =  0. 

8.  (x^/  -  y)dx+(2x'y  +  x)dy  =  0. 

9.  (2xy*-y)dx-[-(xY  +  2x)dy  =  0. 
10    a;t?y      15/- 35a-'' ^ 

ydx'^21y>'-20x' 

Case  VI.    An  equation   in   the   form   -^  ■+■  Fy  —  Q, 

dx 

where  P  and  Q  are  functions  of  x  or  constants,  is  called 
linear.    eJ^*^^  is  an  integrating  factor. 

Example.  -^  —  -  =  xex.  HereP  = ;  \  Pdx  =  —  \ogx  =  \ogx-'^; 

f  dx     X  X  J 

e^  ^=  x-^.  x~^  is  the  integrating  factor.  Multiply  by  this  and 
also  by  dx,  and  the  equation  becomes  x~'^dy  —  x~^ydx  =  e^dx. 
This  is  exact.    Integrating,  x-'^y  =  e^  +  c,  or  y  =  x{e^  +  c). 


FIRST  ORDER  AND  FIRST  DEGREE  11 

EXERCISE  7 

1.  ^  +  ^  =  3x.  3.^  +  ^  =  ^-3. 
ax      X  ax        X 

2.  -^  —  -  =  3  x^.  4.  -^  +  y  tan  x  =  sec  x. 
ax      X  ax 


dy  . 

5.  cos  x-r-  +  7/  =1—  sin  X. 

ax 

6.  cos x[-j-  —  cos x)  =  L—  y 


sin  05. 


7.  xdy  -\- (xy  -}-  y)dx  =  dx. 

8.  xHy  -{•  xy(x  +  2)dx  =  e'^dx. 

9.  xdy  +  ydx  -\-  dy  =  e^(x  -^-l^dx. 

dx       Vx^  + 1 

11.  aj'^c??/  —  ydx  +  dy  =  e^^^~^(x'  +l)dx. 

12.  a;(ar^  +1)(^2/  +  y(2x^  ■^l)dx  =  x(ar*  +1)W 

Case  VII.    If  the  equation  is  of  the  form 

where  P  and  Q  are  functions  of  x  or  constants,  it  can 
be  reduced  to  the  hnear  form  as  follows: 
Multiply  by 

(-w+l)^/-":  (-w  +  l)y';^+P(-w  +  l)«/-'»+i 
=  ^(-^  +  1).  "^^ 

Substitute  ^~'*+i  =  w,  (—n+V)y-'^-f-  =  -—'  The  equa- 
,  dx      dx 

tion  becomes  — +P(— w+1)m  =  ^(— w+1),  which  is 
linear  and  can  be  solved  by  the  preceding  case. 


12  DIFFERENTIAL  EQUATIONS 

EXERCISE  8 

''^  6.  3xff--hx'-2/  =  0. 

^  dx         -^  7.  2xydx  —  2a^dj/  =  i/^efdx. 

8.  /-2x»-3V^  =  0. 

9.  4  y^  -;^  —  3  y*  tan  x  =  3  sec  sc. 

^  dx  ^ 

10.  (2  X  +  3)%  =  4  2/  (2  x  +  3)2(/a;  +  2  y^rZx. 

11.  ^e^dy  =  x«^2^  —  ydx. 


EXERCISE  9 
MISCELLANEOUS  PROBLEMS 

1.  a;(l-logxy)^  +  y  =  0. 

2.  y(l-loga;y)  +  x2  =  0. 

3.  {^x-22y  +  22)dx-^(Xlx-^y-l)dy  =  {i. 

cte      aj''  —  1 

ax      X  dx 

6.  2y^dx  =  x(2y-x^eV)dy. 

7.  xrfy  =  2 ydx  -\-  a^e^dx. 

8.  (e*+ y  +  eJ')  c^y  =  (e*+ !'  +  e^  rfx. 

dx 


FIRST  ORDER  AND  FIRST  DEGREE  13 

10.  y^  =  2xy-x^f- 

11.  (4: X  — 3 i/)dx=(Sx— 10 i/)dy. 

12.  2x^  =  y  +  2  V«. 

ax  *— ^ 

13.  sin  2x  •  dy  =  2  tan y  •  dx. 

14.  3  X  (3  y  c^cc +  4  a;  c??/)  =  10  ?/^(?/(£a;  + 3  ccrfy). 

15.  {x-2y-\-^)dx  +  (Jx  —  14.y-{-l)dy  =  0. 

16.  2a;^  =  y(2- V9/-4x2)^ 
/2a;       1\  ,         /2x^       x\  , 

18.  (ax  -\-  b)dy  =  (by  +  «)  '^a'- 

19.  2xtan-  +  w  =  -— ^• 

ic  ax 

20.  y  (xy  + 1)  rfx  +  x  (xy  —V)dy  =  0. 
21.3  xV%  +  X2/«  (x  +  2)  rfx  =  2  e^dx. 
22.  2  tan  ydx  -\-  tdmxdy  =  0. 

dy  _  sin  y  -\-2x7j  -\-  y  sin  x 
c?x        cos  X  —  x^  —  X  cos  y 

24.  (?/2  +  2y +  4)<Zx+(x2  +  2x  +  4)6?2^  =  0. 

25.  X  (1  —  log  xy)  dy  -\-  y(l  +  log  xy)  dx  =  0. 

26.  (3-5  xV^  (^x  =  2  xV/^2/. 

27.  2sin-^rfx  =  ^^=^^. 

28.  y^c^x  =  2  xydy  +  2  x^fdx. 

29.  6xy%=(l- 3y2  +  5x*)(fo;. 

30.  xy(xdy  +  ycZx)  +  (xc?y  —  yc?x)  Vl  —  x'^  =  0. 


CHAPTER  III 

DIFFERENTIAL  EQUATIONS  OF  THE  FIRST  ORDER  BUT 
NOT  OF  THE  FIRST  DEGREE 

When  the  equation  is  of  the  first  order  but  not  of 
the  first  degree,  -^  will  generally  be  represented  by  p. 

Three  cases  will  be  considered:  (1)  the  equation  solv- 
able for  p ;  (2)  the  equation  solvable  for  y ;  (3)  the 
equation  solvable  for  x. 

Case  I.  Solvable  for  p.  If  the  equation  can  be 
resolved  into  factors  of  the  form 

O  -/i(«»  y)]  Ip  -M^^  y)]  etc., 

each  factor  may  be  put  equal  to  0  and  these  equations 
solved  by  the  preceding  chapter. 

Example,  p'^  —  2 p  —  B5  =  0.   This  can  be  factored  into 
(p  +  5)(p-7)=0. 

Placing  each  factor  in  turn  equal  to  0,  and  solving,  we  have 
y  +  5x  +  c  =  0  and  y  —  7 x  +  c  =  0. 

Since  the  differential  equation  is  of  the  first  order,  the 
solution  can  contain  but  one  arbitrary  constant.  Hence 
the  c  must  be  the  same  in  eaeh  separate  part  of  the 
complete  solution.  This  solution  in  the  given  example 
may  also  be  written  (y  +  5  a;  -f  c)  (y  —  7  a;  +  c)  =  0. 

14 


DIFFERENTIAL  EQUATIONS  OF  FIRST  ORDER     15 

EXERCISE  10 

1.  y  -  6j9  +  8  =  0.  4.  If  -  3/  -  3j9  +  2  =  0. 

2.  f-x>-Vl  =  ^.  5.  i^  -  Ojs^  +  26^  -  24  =  0. 

3.  f  —  f  —Vljp  =  0.        6.  p^  +px  +  py  +  ajy  =  0. 

7.  y  -  6/a;  -^llpx"  -  6a;'  =  0. 

8.  ^Xa;+l)+y=p(x  +  y)  +  l. 
9.  xy  -l=x'p-p.      11.  yo;'  -  /  =  0. 

10.  ajy  +  a;^  =  j9^.  12.  a;^^  +p{x^  +  y^)  + xy  =  0. 

Case  II.  If  the  equation  can  be  solved  for  y,  do 
so,  and  then  differentiate  with  respect  to  x^  remember- 
ing that  -^=p.    If  possible,  solve  this  equation  by  the 

second   chapter.     By    eliminating   the  p   between    this 
result  and  the  original  we  obtain  the  required  solution. 

Example,  p^x  =  2py  +  x. 

p^x  —  X 
Solving  for  y,  y  =      gj,      " 

Differentiating  with  respect  to  x, 

p(2px-j-  +  p^  —  1)  —  (p^x  —  x)-^ 
ax  ax 

^  = 27^ 

Clearing  of  fractions  and  collecting  terms, 

x(p^  +  l)'f-=p(p^ +  1),  or  x^-^=p. 
ax  ax 

X 

The  solution  for  this  equation  is  p  =  -• 

c 

Eliminating  the  p  between  this  and  the  original  equation, 
a;2  =  2cy  +  c\ 


16  DIFFERENTIAL  EQUATIONS 

The  equation  y  =px  +/( p)  is  known  as  ClairauCs 
equation.     It   comes    under    this    case.     Differentiating 

with  respect  to  Xy  we  have  p=p  +  x^-\-f'(p^-^  or 
,  ax  ax 

-f-  \^-\-f(^p)\  =  0.    Placing  the  first  factor  equal  to  0, 

^^  dp 

we  have  -^=  0,  or  p  =  c.    Hence  the  solution  of  the 
ax 

differential  equation  is  t/  =  cx  -\-f(^c). 

If  the  second  factor  is  placed  equal  to  0,  and  the  p 
eliminated  between  this  and  the  differential  equation, 
the  result  will  be  a  solution  satisfying  the  differential 
equation.  It  will  contain  no  constant  of  integration, 
and  hence  is  not  the  complete  solution.  Neither  can 
it  be  obtained  from  the  complete  solution  by  giving 
a  particular  value  to  the  constant.  It  is  called  the 
singular  solution,  and  is,  generally,  the  envelope  of  the 
family  of  lines  represented  by  the  complete  solution. 

EXERCISE   II 

1.  y  =  2px  —  p^.  4.  4x^7/  =  22>x^  -\-  p^- 

2.  y  =  p  -\- p^e~^^.  5.   Qpx^ -\- A^p"^  = 'dxy. 

3.  y  =px  logx  +p^x^.  6.  p^x  =  3p^y  -\-  X. 

The  following  are  types  of  Clairaufs  equation.  Solve 
and  obtain  the  singular  solution  if  possible. 

7.  y  =px  +p^.  10.  py  =p^x  +  a. 

8.  y  =  px -\- t3in~'^p.  11.  y  =  px  +  p  ^/ -\-l. 

9.  y=px-\-Q^p'^+\.         12.  f—l+p'^{x^ -1)=2 pxy. 

Case  III.    If  the  equation  is  solvable  for  x,  differ- 

dx      1 
entiate  with  respect  to  y,  remembering  that  -—  =  — 

t/  IT 


DIFFERENTIAL  EQUATIONS  OF  FIRST  ORDER     17 

Solve  this  equation,  if  possible,  by  the  preceding  chapter, 
and  eliminate  the  p  between  the  resultant  and  the  origi- 
nal equation. 

Example,    y  —  ^px  +  phj. 

II  —  p*v 
Solving  for  x,  x  =  - — ^— ^  • 

'dp 


Differentiating  with  respect  to  y, 

dp 

dyl     '■"      ^    "''dy 


py\.-p^^-'-p^y-^\-iy-p^y)Y 


p  3p^ 

Clearing  of  fractions  and  collecting  terms, 

y^(pi  +  2)  +  2p(p^  +  2)  =  0, 
ay 

or  i^=-2^. 

p  y 

Solving,  p  =  ^- 

Eliminating  the  p  by  substituting  in  the  original  equation, 
/  =  3  A  +  c3. 

EXERCISE  12 

1.  y  =  2px  -{-  4:p^y.  6.  x  —  y=p^* 

2.  x-2p  +  Zp"  =  0.  7.  p^x  =p  +  e-2y. 

3.  y  =  2px  +  y^(2p)^.  8.  9xyp''  —  6  fp  -\- 4:  =  0. 

4.  2p^x  -  3p^y  =  2y.  9.  p'^e'^v  =  pe^v  +  e". 

5.  p^^y  +  2px  =  1.  10.  A.p'^xf  =  2p'if  + 1. 

*  When  the  elimination  of  p  is  impossible,  the  solution  is  indicated 
by  obtaining  x  and  y  in  terms  of  p. 


18  DIFFERENTIAL  EQUATIONS 

EXERCISE  13 
MISCELLANEOUS  PROBLEMS 

1.  fo?'  =  2xy  +  x^p'. 

2.  ey=peP^. 

3.  phc=py\o^y-\-f. 

4.  fx"  +if  =  2p{xy-{-2 a*). 

5.  xy(p^-\-l)  =  p(x^-hf). 

6.  p^x"  -  a"^  +  9/ =  b"" -h  2pxy. 

8.  2pxy  =  xy  +  f-a^(j>^+V). 

9.  2  x^y  =  3  x*p  -\-  y*p*.  (Substitute  2^  =  v  and  solve  for  v.) 

10.  y^  -hy*  +  x'p^  =  2  xyp  +  2  x^^. 

11.  x(jy'-\-2)=p(2x'+l). 

12.  2/(1— 4a;j9^  =  2;j(a;  — 2/=^  +  l).  (Substitute  jT*  =  y  and 
solve  for  v.) 

13.  4  x2+ 13  a;jrp  + 9  2/^i>'  =  0. 

14.  7/(j)^  +  l)  =  ^(x  +  ypf.    (Substitute  4 ar^  +  3 ^  =  v».) 

15.  4  £c  +  9  J59^  =  6p  {x  +  y). 

17.  xp^(x-\-y)=f(^—p). 

18.  2x22y'»+5a-y^>  +  2/  =  0. 

19.  ar*2/  +  yP^  —  ^^P- 

20.  7f={e-^^-l){x?  +  2py). 

21.  p{x^  -  'f)-{-  xyif  -1)=  a'p.    ^Substitute  ^  ^  ") 


CHAPTER  IV 

DIFFERENTIAL  EQUATIONS  OF  ORDERS  HIGHER  THAN 
THE  FIRST 

If  the  dependent  variable  and  its  derivatives  are  all 
of  the  first  degree  and  are  not  multiplied  together,  the 
equation  is  called  linear.  Equations  of  this  kind  can  all 
be  put  into  the  general  form 

-^+A T+^S 


A  '±J.+A.- I4.A  - I,.. A      v-fCx^ 


where  A^,  A^,  A^,  etc.  are  functions  of  x  or  con- 
stants. We  shall  consider  first  the  equation  all  of 
whose  coefficients  are  constants  and  whose  second 
member   is    0. 

To  solve,  substitute  1/  =  e^^.  This  will  lead  to  an 
auxiliary  equation  in  k.  Obtain  the  roots  of  this  equa- 
tion, Jc^,  Jc^,  etc.  The  differential  equation  is  satisfied  by 
any ,  of  these  values  in  y  =  e^'^  or  by  the  combined  form 
^=Cje*i^+C2e*«^+  etc.   The  latter  is  the  complete  solution. 

Example.   ^-2^-8y  =  0. 
dx^         dx 


dx^         dx 

Substitute  y  =  e^^;  then  —  =  ke^^  and  — ^  =  ^%*^.    These,  in 
dx  dx^ 

the  differential  equation,  give  e*^^  (k^—2  k—8)  =  0,  or  P— 2  k—  8  =  0. 
The  roots  of  this  equation  are  4  and  —  2,  and  the  complete  solu- 
tion is  ^  =  c-^e^^  +  c^e-^^. 

19/  ^ 


20  DIFFERENTIAL  EQUATIONS 

Three  cases  are  to  be  considered :  (1)  when  the  roots 
of  the  auxihary  equation  are  all  real  and  different; 
(2)  when  some  or  all  of  the  roots  are  complex ;  (3)  when 
some  of  the  roots  are  repeated. 

Case  I.    This  case  is  solved  like  the  problem  given. 


EXERCISE 

14 

1. 

d-'y 
dx" 

-7|-8.  =  0. 

^-   dx'  ~ 

-25y 

2. 

d'y 
.da? 

-6|  +  8.  =  0. 
daf         dx 

0. 

4     ^^V_ 
da? 

dx 

10.  ^  +  482/ =  28^. 
dx^  ^  dx 

Case  II.  Some  of  the  roots  complex.  The  complex 
roots  may  be  written  in  the  form  a  ±b  "v^,  and  these 
occur  as  conjugate  pairs.  The  terms  corresponding 
would  be 


ORDERS  HIGHER  THAN  THE  FIRST  21 

The  quantity  in  the  parentheses  may  also  be  written 
Cj  cos  hx  +  CjV— 1  sin  hx  +  c^  cos  hx  —  CgV— 1  sin  6a;, 
or         (cj  +  Cg)  cos  hx  +  (^i"^^l  —  <?2  '^~1)  sin  Ja;, 
or,  again,  Ar^  cos  62;  +  ^2  sin  hx. 

The  terms  then  become  e"^(^j  cos  Ja;  +  h^  sin  6a;). 

,    Example.   — \  +  2—  +  4v  =  0.     The    auxiliary    equation    is 

^^^+2^  +  4  =  0,  of  which  the  roots  are  —  1  +  V—  3  and 
—  1  —  V—  3.  Here  a  =  —  1  and  h  =  V3,  and  the  complete  solu- 
tion is  y  —  e-^ (cj cos  v3  x  +  Cg  sin  v3  x). 

EXERCISE   15 


1. 

S-«S+"-»- 

^•S-3— 0- 

2. 

2-«t+^«^=«- 

«-S+l+^=«- 

3. 

2+*^=«- 

^■S-^- 

4. 

3  +  .V  =  o. 

s.g-ie.  =  o. 

-«S-^«S+«» 

(Za;2 

-50|+12,=  0. 

If  the  operator  — -  be  replaced  by  D,  -r-z  by  i)^,  etc., 
,,  ,.  ax  dx^ 

the  equation 

dx'         dx         ^ 
may  be  written 

IJf^y-^Dy-^y=0,     or     (Z>2- 2i)- 8)2/  =  0. 


22  DIFFERENTIAL  EQUATIONS 

The  coefficient  of  y  is  the  same  function  of  2>  as 
the  auxihary  equation  is  of  k.  If  this  coefficient  is 
factored  as  an  ordinary  quantic,  the  equation  becomes 

(i>_4)(i)+2)y=0. 
If  the  y  is  operated  on  by  the  near  factor  {B  +  2),  the 
result  is  -^  +  2  y ;  operating  on   this  result  with  the 
other  factor  (J)  —  4),  we  obtain 

aar         ax         ax  dor         ax 

the  original  differential  expression.  It  can  easily  be 
shown  that  the  order  of  the  factors  may  be  changed 
without  altering  the  result. 

Case  III.  When  the  auxiliary  equation  has  multiple 
roots. 

If  two  of  the  roots,  k-^,  k^,  of  the  auxiliary  equation 
are  equal,  we  cannot  use  Cje*>*+  Cgg*'*,  as  this  could 
be  written  (c^  +  c^e^^'^^  and  the  c^  +  ^2  ^^  equivalent  to 
one  constant.  The  way  to  handle  this  case  can  best  be 
illustrated  by  an  example. 

The  roots  of  the  auxiliary  equation  are  3,  3.    Using  the 

D   symbol,   the   differential    equation    may   be   written 

(i)2_6i>+9)y  =  0,     or     (2)- 3)(i)- 3)y  =  0. 

Placing  (D  -  3)  y  =  V,  (1) 

the  equation  becomes 

(D-3)v  =  0,     or     ^-3t;  =  0. 
ax 


ORDERS  HIGHER  THAN  THE  FIRST  23 

This  equation  is  linear  with  e~^^  as   the   integrating 

factor.   Its  solution  is  t?  =  c^^^.    Substituting  this  in  (1), 

we  have  ^ 

(I)-Z')t/  =  c^e^=',     or     -^-^y=c^^''. 

This  equation  is  again  linear  with  e"^*  as  the  integrat- 
ing factor.  The  solution  is  y  =  (c^x  +  c^  e^^  which  is 
the  complete  solution  of  the  original  equation.  In  the 
same  way,  if  three  of  the  roots  of  the  auxiliary  equa- 
tion were  equal,  y=^''(c^a^+c^x  +  c^  would  be  the 
corresponding  part  of  the  complete  solution,  etc. 

EXERCISE  16 

dx*  dx^  dx^ 

6.^-5^  =  0. 
dx^         dx* 

dx*  dx^  dx^  dx  "^ 

7.  (4  2)2- 4  Z) +1)2/ =  0. 

8.  (D^  +  5  ly  +  3  D  -  9)?/  =  0. 

9.  (7)3  +  7)2  _  5  /)  _|_  3)y  =  0. 
10.  (7)^-6  7)2+12  7)- 8)y  =  0. 


24  DIFFERENTIAL  EQUATIONS 

Case  IV.  When  the  second  member  is  not  0,  the 
complete  solution  will  consist  of  two  parts :  the  general 
part,  which  contains  the  constants  of  integration ;  and  the 
particular  part,  which  accounts  for  the  second  member. 
The  complete  solution  is  the  sum  of  the  two  parts.  The 
general  part  is  obtained  by  one  of  the  three  preceding 
cases ;  the  particular  part  depends  upon  the  second 
member,  and  various  methods  might  be  given  for  ob- 
taining it.  We  shall  confine  ourselves  to  a  second 
member  comprising  terms  of  the  forms  aj",  e"%  sin  wo;, 
cos  Tix,  and  their  products,  and  shall  use  an  inspection 
method  to  arrive  at  a  corresponding  solution. 

Example  1.  — ^ ^  —  6w  =  3x2  —  ox  +  6.  Here  the  auxiliary 

dx^      ax 

equation  is  P  —  i  —  6  =  0,  its  roots  are  3  and  —  2,  and  the  general 
part  of  the  solution  is  ?/  =  c^e*^  +  Cge"^^.  The  second  member  of 
the  given  equation  must  have  come  from  a  power  series  of  the 
form  c^x^  +  c^x  +  Cy  Substituting  y  =  c^x^  +  c^x  +  Cg  in  the  given 
equation,  we  have 

2  Cg  —  2  c^x  —  c^  —  6  CgX^  —  6  c^x  —  6  Cg  =  3  z^  —  5  a;  +  6. 
Equating  the  coeflBcients  of  like  powers  of  x, 

—  6  Cj  =  3 ;  —  2  Cg  —  6  c^  =  —  5 ;  2  Cg  —  c^  —  6  Cj  >=  6. 
Solving  these  equations, 

x^  4 

Hence  the  particular  part  of  the  solution  is  —  77  +  ^  ~"  5- '  and 

x2  4 

the  complete  solution  is  y  =  Cjg**  +  c^e-^^  ~  o"  "^  ^  ~  q  ' 

Example  2.   — |  —  5-^  +  4m  =  2€'^.    Here  the  auxiliary  equa- 
dx^        dx 

tion  is  i^  —  5  ^'  +  4  =  0  and  the  general  part  of  the  solution  is 
y  =  Cj«*^  +  c^e"^.  The  second  member  of  the  given  equation,  2  e**. 


ORDERS  HIGHER  THAN  THE  FIRST  25 

must  have  come  from  a  term  of  the  form  ce^^.  Substituting 
y  =.  ce^x  in  the  given  equation, 

9  ce^^  - 15  ce8«'  +  4  ce^^  =  2  e^^,     or     -  2  c  =  2. 

.-.  c=-l. 

The  particular  part  of  the  solution  is,  then,  —  e*^,  and  the 
complete  solution  is  ?/  =  c^e*-^  +  CgC-^  —  e^^. 

Example  3.    — ^  —  5  -^  +  6  w  =  sin  x.    Here  the  general  part  of 
(/x''         ax 

the  solution  is  _y  =  c^e^*  +  CgC^^.  The  second  member,  sin  x,  must 
have  come  from  an  expression  of  the  form  Cg  sin  x  +  c^  cos  x. 
Substituting  y  =  c^  sin  x  +  c^  cos  z  in  the  given  equation, 

(5  Cg  +  5  c^)  sin  a;  +  (5  c^  —  5  Cg)  cos  a;  =  sin  x. 

Equating  corresponding  coefficients, 

5  Cg  +  5  c^  =  1 ;     5  c^  —  5  Cg  =  0. 

Solving  these  equations,  Cg  =  c^  =  y"^. 

Hence  the  complete  solution  is 

y  -  Cje^^  +  CjC*^  + — 

Example  4.   — ^-\-  2  —  =  e~^^.    Here  the  general  part  of  the 
dx^         dx 

solution  is  y  =  c^-\-  c^e~'^^.    It  might   appear  that  the    second 

member  must  have  come  from  a  term  of  the  form  ce-^'^;  but 

this  is  already  found  in  the  general  part  of  the  solution.    In  this 

case  assume  y  =  cxe-'^^.    Substituting  in  the  given  equation, 

4  cxe-2^  -  4  ce-^^  +  2  ce-2«  —  4  cxe-^^  =  e-^^^. 

•••  -2cg  =  1,     or     Cg  =-  J. 

Hence  the  complete  solution  is 

y  =  Cj  +  CjC"'^^  —  \xe-^^. 

If  the  second  member  consists  of  several  terms,  the 
corresponding  particular  integral  may  be  found  for  the 
parts  taken  separately. 


26'  DIFFERENTIAL  EQUATIONS 

EXERCISE  17 

1.  (D»-3i>2  +  3Z)-l)y  =  e='^. 

2.  (Z>8-3Z)^+3Z>-9)y  =  7e2^. 

3.  (Z>8-6i>2  +  llZ>-6)?/  =  e**.     ' 

4.  -rr  —  Id  7/  =  sm  x. 

5.  {Lf-12D-^21)y=^lx-  171. 

7.  (Z>2-l)y  =  3a;-2sr«. 

8.  (Z>*  +  4)?/  =  cos2a;. 

9.  (i)2-Z)-12)y=  e*^  +  sin3a;  +  e2^sin3x. 
10.  (6D2-7Z)4-2)2/  =  e^. 

Case  V.  When  the  equation  is  of  the  form 

A^y  A^y  etc.  being  constants,  substitute  x  =  eK    Then 

di 
dy  _dy  dt  _  dt 
dx      dt     dx      e' 

dy      dy 
or  x-^  =  -^' 

dx      dt 


ORDERS  HIGHER  THAN  THE  FIRST  27 

Using  the  symbol  D  for  —  >  we  have 

=  D  (I>  - 1)  (D  -  2) «/,  etc. 

This  substitution  will  reduce  the  equation  to  one  of  the 
preceding  forms.    In  the  same  way,  if  the  equation  is  of 

the  formal  Cax+  by^=^  +^„(aa;+6)"-i  - — ^  +  . .  .  =  0, 

substitute  ax  +  b  =  e^ 

.  EXERCISE  18 

daf  dx       ^ 

2.  ^_^^4.il  =  l 
da^      X  dx       x^ 

dh^k 


28  DIFFERENTIAL  EQUATIONS 

10.(.-l)3g+(.-l)^g4-3(x-l)|-8,  =  0. 

Case  VI.    If  the  equation  does  not  contain  y  directly, 
represent  the  derivative  of  lowest  order  by  p.    Then  the 

next  derivative  will  be  -^  >  etc.    Solve  this  new  equation 
ax 

for  p  in  terms  of  x,  and  ultimately  for  i/. 

Putting  -^  =  »,  X  -p-  =  »  +  »*. 

ax  ax  . 

Separating  the  variables, 

dp     _  dx 
p^  +  p       X 
dp        pdp    _  dx 
p       p^  +  1        X 

j5  ,   .        ,  dy  dy  ±x 

Solving  for  p  =  -f-,  -j- =  ; 

dx  dx       Vc^  — x^ 


whence  ^  =  ±  "V  cf  —  3^+  c^. 

Clearing  of  radicals, 

x^ + (y  -  c^r  =  c^. 

EXERCISE  19 
dx^       dx  dx^      \dx/ 


ORDERS  HIGHER  THAN  THE  FIRST  29 


5.  ^-4  =  2  tan  X  -^r 
dx^  dx\' 


\dx^)      \dx/         dx 


7..(.-l)g+(2x»-l)|  =  0. 

8.  -r^  =  (2  tan  X  -f  cot  cc)  ^  • 
o^x^      ^  ^  dx 


9    w^-l-^. 


2nl 


[-©1 


10.  — =  a. 

d}y 

da? 

Case  VII.  If  the  equation  does  not  contain  x  directly, 

,  dy  ^1       dP'v     dp    dy        dp 

put  -^=p\  then  -^  =  -J-.-f-=p-f- 
.      dx  dsr      dy    dx         dy 

Example.   (l-,.)g  +  ,(|)'=0. 

Putting  |!  =  i»,       (l-j,>^+p=y  =  0. 

Separating  the  variables, 

dp  _  ydy 

j-~  (i-y-^y 

Solving,  p  =  c^Vl  -y^^-^'-> 

whence  sin-^y  =  c^x  +  c^. 


00  DIFFERENTIAL  EQUATIONS 

EXERCISE  20 

Case  VIII.    When    the    equation    is    of    the    form 
— ^  =/(y),  multiply  by   1-—  dx   and   integrate.    This ' 
gives  I  -^ )  =  2  I  /(^)  c?y  +  c-    Extract  the  square  root, 
separate  the  variables,  and  solve. 

Example.   ^  =  y- 

Integrating,  \-£\  =  ^H  c^, 

or  -/  =  ±V/  +  c,; 

ax 

V2^2  +  Cj 


ORDERS  HIGHER  THAN  THE  FIRST  31 

Using  the  positive  sign, 


log  (y  +  Vy  +  q)  =  X  +  log  C2, 
y  +  V^2  ^  (.^  —  c^e*^. 

Clearing  of  radicals,  ^2+  Cj=  i/^  —  2  c^ye'  +  c^e^^. 


Solving  for  y. 

^        2        2c2 
EXERCISE  21 

> 

-Z-'y 

-S=2^-     '• 

-  S=-«v- 

6     "^'^Z.    1                   8 

dx^          -y/y 

y  dx" 

-^S=i- 

e.^g  =  l.            . 

y  dx^    • 

10. 

8    dhj        . 

Case  IX.  An  exact  differential  equation  has  been 
defined  as  one  obtained  from  its  primitive  by  differen- 
tiation without  further  reduction.  If  the  equation  is 
of  an  order  higher  than  the  first,  and  is  of  the  form 

^^1^'^^'^  ~1^  ^"'  ^'^+^^  =-^(^)'  "^^^^^  ^1'  ^2'  etc. 
are  functions  of  x,  the  test  for  exactness  is  as  follows: 
Differentiate  A-^  with  respect  to  x  and  subtract  the 
result  from  A^ ;  differentiate  this  remainder  with  respect 
to  X  and  subtract  from  ^3,  etc.  If  the  last  remainder 
is  0,  the  equation  is  exact.    If  the  test  is  satisfied,  the 


32  DIFFERENTIAL  EQUATIONS 

first  integral  is  obtained  as  follows :  The  first  term  will 

be  A.  — — ^ ;  the  coefficient  of  the  second  term  will  be 

the  first  remainder  obtained  in  the  test ;  the  coefficient 
of  the  next  term  the  next  remainder,  and  so  on.    The 

second  member  will  be  /  f(^x)  dx  +  c.    If  the  resulting 

equation  is  again  exact,  repeat  the  process.    Otherwise, 
solve,  if  possible,  by  one  of  the  preceding  cases. 

Example.    (3  a;2  -  5  x)  ^  +  (12  X  -  10)^  +  6  V  =  0 

6x-    5  ^^      6 

FoUowine  the  test,  - 

^  6x-    5  0 

Since  the  last  remainder  is  0,  the  test  is  satisfied.   The  first 
integp'al  is  , 

(3  x2  -  5  or)  ^  +  (6  X  -  5)y  =  Ci 
'^^        6x-5. 
0 

The  test  is  again  satisfied,  and  the  complete  solution  is 

(3  x^  —  5  x)  2^  =  CjX  +  Cg. 

EXERCISE  22 
M^-3.)g.-(6.-9)g^6|  =  .  +  6. 

"2:>(2a;  +  5)^g  +  8(2x  +  5)g  +  8y  =  sina:. 

o    ■     ^V  .  o       ^y        ■         f 

3.  sm  X  -r-^  +  2  cos  x-r  —  V  sm  x  =  e*. 
da^  dx      ^ 


ORDERS  HIGHER  THAN  THE  FIRST  33 

5.  (3x  -  4.fPi  +  6(3x  -  4)^  =  0. 

6.  (3x^-5x+7)g+(12x-10)g  +  6y  =  e-. 

7.  ic^^  +  21x«^  +126a;^^  +  210icV  =  3  e'^  +  sin  a;. 

(Zx^  dx^  dx 

8.  — ^  =  sinx  +  8e2^. 

d^y      d^y 
c?ic^      dx^ 

10,  (a^-x'')ip{  =  ax. 
^--^  dx^ 

Case  X.  In  Case  V  a  change  of  the  independent 
variable  reduced  the  equation  to  a  solvable  form.  In 
many  other  problems  a  proper  change  of  either  the 
dependent  or  independent  variable  will  effect  a  reduc- 
tion of  the  given  equation.  This  is  especially  true  of 
equations  of  the  second  order.  We  shall  consider  two 
cases,  and  shall  illustrate  the  methods  by  examples. 

Example.    — ^  —  2  tan  x  —  —  y  =  0. 

dx^  dx 

T-  ^^.  dy         dv   ,      du    d^y         dHi   ,  ^du    dv   ,      dht 

Putting  y  =  uv,-^  =  u \-  v  — ,  — =^  =  u  — -  +  2 V  v  — - 

dx         dx        dx    dx^         dx^         dx    dx         dx^ 

in  the  given  equation  and  collecting  terms, 

« — s— (2 2  tanar) f-( —  2  tanw  —  u)v  =  0.      (1) 

dx^      \    dx  I  dx      \dx^  dx    / 

Placing  the  coefficient  of  —  equal  to  0,  and  solving  without 
constants  of  integration, 


34  DIFFERENTIAL  EQUATIONS 

Then  —  =  sec  x  tan  x, 

ax 

and  — -  =  sec' a;  +  sec  x  tan*  a:. 

Substituting  in  equation  (1)  and  simplifying, 

^  =  0; 

whence  v  —  c-^x  +  c^. 

Therefore  y  =  uv  =  (cyX  +  c^)  sec  x. 


EXERCISE  23 
1.  0  -  2  cotcc^  +  y{2  cot^x  +  6)  =  0. 

3.  ^  -  (1  +  2  cotx) ^  +  2/(1  +  cotic  +  2  cotV)  =  0. 

5.  4x»^-20x^  +  y(35  +  36a:2)=0. 

6.  -r^  4-  2  a  cot  aa;  -;^  -  2  a^y  =  Q. 

8.  :r>g-2(ar'  +  a;)g+y(5x»  +  2:.  +  2)  =  0. 

9.  (x\osxf^-'ix\oex^-yi(x\ogxf-\ogx-21=a. 


10.^I„g.(g  +  av)=.-2.|. 


ORDERS  HIGHER  THAN  THE  FIRST  35 

Case  XI.    In  this  case  the  equation  is  reduced  by 
changuig  the  independent  variable. 


Example.    9x^  +  6^  +  4  x'^^^  =  0. 
dsr         ax 


^d  b: 


Putting  X  =f(t),  .K: 

^.  dy      dy    dt 

then  -2.  =  ^.—, 

dx      dt    dx  T^j_-       .V 

and  ^  =  ^/^y+^.i^^       A'   '^^ 

dx''    df^ydx)    dt  dx^  ^ 

Substituting  in  the  given  equation  and  collecting  terms, 

„     d'^yldtS^  ^  dy(^     dH   ,  ^dt\  ,    .    ^i        _ 

dt^  \dxj       dt\      dx^        dx)  ^  ^  ^ 

dy 
Placing  the  coefficient  of  -^  equal  to  0  and  solving, 

dt 

<  =  3  xi 
Substituting  in  (1)  and  reducing, 

dt^         •' 
The  solution  of  this  equation  is 

y  =  c^  cos  f  <  +  ^2  ^^''^  %  '• 
Hence  y  =  ^i  cos  2  x^  +  Cg  sin  2  a;». 

This  result  may  also  be  written 

y  =■  c-^  sin  (2  a;'  +  *)• 

Placing  the  coefficient  of  -^  equal  to  the  coefficient 
of  ^  in  (1)  will  often  effect  an  easy  reduction. 


36  DIFFERENTIAL  EQUATIONS 

EXERCISE  24 

1.  — |  —  tana;^  +  4y  sec^a;  =  0. 
dx^  ax 

d^y         2x    dy  y        ^ 

4.  ^-2csc2a;^-4ytan2a;  =  0. 
oar  ax 

7.  ^  +  2csc2a;^-2/cot=^a;  =  0. 

dx^      dx  ^ 

d^y      ,         dy  on. 

9.  -r^  —  tan x—  —  y sec-a;  =  0. 

EXERCISE  25 
MISCELLANEOUS  PROBLEMS 

3.  (Z>*  - 10  Z)8  +  25  D^  y  =  0. 

6.  ^  —  (cotx  +  4tana;)-^  — 4ytan^x  =  0. 

(Substitute  s  =  log  sec  x.) 


ORDERS  HIGHER  THAN  THE  FIRST  37 

11.  (2x-3)^g  +  2(2x-3)|-4aV  =  0. 

12.  g+(tanx-2)g  +  y(l-tanx)=0. 

(Substitute  y  =  ve*.) 

13.  (Z>«-2D2-15D)3/  =  0. 

14.  — |  =  2tanx-/-- 

15.  ^  +(tana;  -  2  cotx)  ^  +  2 2/  eot^x  =  0. 

(Substitute  «  =  log  sin  x.) 


dx      dx^      \dx) 


17.  (a;  logxf  ;r|  +  x  logx(loga;  —  4)^  +  6?/  =  0. 

(Substitute  z  =  log  log  x.) 

18.  (D*  +  4)?j  =  0.  ^  &     6     / 

19.  (i)2-lli)+30)2/  =  e«^-e«^. 

rfa:^      \c?x/ 

21.(^+l)S-2^(|)=(^+l)'<--V. 

22.  ^  =-»'.. 

23.  (ii*-10B'  +  50I)»-130/)  +  169)3/  =  0. 


88  DIFFERENTIAL  EQUATIONS 

26.  (II^  —  AlJ^  +  4:D)y  =  0. 

dx^       \ax/  ax 


EXERCISE  26 
APPLICATIONS 

1.  Find  the  equation  of  the  curve  whose  subnormal  is 
constant. 

2.  Find  the  equation  of  the  curve  whose  subnormal  is 
proportional  to  the  square  of  the  abscissa  of  the  point  of 
contact. 

3.  Find  the  equation  of  the  curve  whose  subnormal  is 
equal  to  the  abscissa  of  the  point  of  contact, 

4.  Find  the  equation  of  the  curve  in  which  the  slope  at 
any  point  varies  directly  as  the  abscissa  of  the  point. 

5.  Find  the  equation  of  the  curve  in  which  the  slope  at 
any  point  varies  inversely  as  the  ordinate  of  the  point. 

6.  Find  the  equation  of  the  curve  in  which  the  slope  at 
any  point  varies  directly  as  the  ordinate  of  the  point. 

7.  Find  the  equation  of  the  curve  whose  slope  at  any 

4a; 
point  is  —  - — 
9y 


APPLICATIONS  39 

8.  A  point  moves  in  a  path  always  perpendicular  to  the 
line  joining  its  position  to  the  origin.  Find  the  equation  of 
its  path. 

9.  The  tangent  to  a  curve  cuts  intercepts  from  the 
coordinate  axes  whose  sum  is  constant.  Find  the  equation 
of  the  curve.  (Singular  solution.) 

10.  The  tangent  to  a  curve  cuts  intercepts  from  the 
coordinate  axes  whose  product  is  constant.  Find  the  equa- 
tion of  the  curve.  (Singular  solution.) 

11.  Find  the  equation  of  the  family  of  curves  all  of  which 
cut  the  hyperbola  3?  —  if  =  d^  at  right  angles. 

12.  Find  the  equation  of  the  family  of  lines  all  of  which 
cut  the  circle  ar*  -f-  ^  =  a**  at  right  angles. 

13.  A  point  moves  so  that  its  acceleration  varies  inversely 
as  the  cube  of  its  distance  from  the  initial  point.  Find  the 
equation  of  its  motion. 

14.  A  point  moves  so  that  its  acceleration  varies  inversely 
as  the  square  of  its  distance  from  the  point  of  starting.  Find 
the  equation  of  its  motion, 

15.  A  point  moves  so  that  its  acceleration  varies  directly 
as  the  distance  from  the  initial  point  and  is  negative.  Find 
the  equation  of  its  motion. 

16.  Find  the  equation  of  the  curve  whose  radius  of  curva^ 
ture  is  constant. 

17.  Find  the  equation  of  the  curve  whose  radius  of  curva- 
ture is  twice  the  length  of  the  normal. 

18.  Find  the  equation  of  the  curve  in  which  the  radius 
of  curvature  at  any  point  varies  directly  as  the  slope  at 
the  point. 


40  DIFFERENTIAL  EQUATIONS 

19.  If  a  horizontal  beam  is  supported  at  both  ends  and 
a  load  is  uniformly  distributed  along  its  length,  the  upper 
fibers  are  in  compression  and  the  lower  in  tension.  Between 
the  two  is  a  neutral  surface,  and  its  intersection  with  a 
vertical  plane,  parallel  to  the  axis  of  the  beam,  is  called  the 
neutral  line.    Its  differential  equation  is 


-S=l(?-4 


the  origin  being  at  the  middle  of  the  beam,  x  horizontal, 
and  y  vertical,  I  the  length  of  the  beam,  w  the  weight  per 
unit  of  length,  E  the  modulus  of  elasticity  of  the  beam,  /  the 
moment  of  inertia.    Find  the  equation  of  the  neutral  line,  if 

//•J/  7 

-T^  ==  0  when  ic  =  0,  and  y  =  0  when  x  —  -- 
(toe  ^t 

20.  If  a  load  P  is  concentrated  at  the  center  of  the 
beam  and  the  weight  of  the  beam  neglected,  the  differential 
equation  of  the  neutral  line  is 


Solve  the  equation  under  the  same  conditions  as  above. 

21.  A  cantilever  beam  has  one  end  unsupported.  When 
uniformly  loaded,  the  differential  equation  of  the  neutral 
line  is 

the  origin  being  at  the  fixed  end.    Solve  the  equation  under 
the  conditions,  when  a;  =  0,  ?/  =  0,  -^  =  0. 


APPLICATIONS  41 

22.  If  the  weight  P  is  concentrated  at  the  free  end  and 
the  weight  of  the  beam  neglected,  the  differential  equation 
of  the  neutral  line  is 

Solve  under  the  same  conditions  as  above. 

23.  If  R  is  the  resistance  of  an  electrical  current,  C  is 
the  current,  V  the  electromotive  force,  L  the  coefficient  of 

dC 
self-induction,  L— — \-  RC  —  V.    Find  C  in  terms  of  t. 
at 

24.  The  equation  for  the  current  C  in  a  circuit  consist- 
ing of  a  charged  condenser  and  a  resistance  R  is 

dt  ^  RK       R^  ^^' 

where  K  is  the   capacity  of  the   condenser.    Find   C   in 
terms  of  t. 

26.  The  equation  for  the  quantity  of  electricity  dis- 
charged by  a  condenser  in  a  circtdt  consisting  of  a  condenser 

of  capacity   K  and  a  resistance   i?   is   -^  -|-  -7-  =  —  f(t). 
^.    .        .   \  p  dt      RK      R-'  ^  ^ 

Jbind  q  m  terms  01  t. 


ANSWERS 


EXERCISE  1 


l.x^  +  2xy^/  =  y^.  6.  ^  -  off  +  11^-6,  =  0. 

dx  dx^        dx^  dx 

"dx  \dxj    ^  dx2^W/ 


-KI)'-'(--l)' 


8.^+642^  =  0. 
dx^ 


4.f|-7^  +  10,  =  0.  9.^-tanx^-,sec^x  =  0. 

dx^        dx  dx^              dx 

6.x^  +  f^  =  0.  10.(l-x^)^-x^  +  4,  =  0. 

dx^      dx  dx^        dx 


-■Mt)l=^iW- 


EXERCISE  2 

1.  x2  =  ce-v.  4.  tan^x  +  tan^^/  =  c.      7.  tan x  •  tan  j/  =  c. 

2.  j/  =  ce-^,  5.  X2/  =  c.^ 8.  |/(e^— 1)  =  c. 

8.  tanx  +  tani/  =  c.      6.  y  =  cVx^  —  l.  9.  secx- secy  =  c. 

10.  tan X  +  tan ?/  =  x  +  c.  14.  y  =  c (x^  +  x  +  1). 

11.  X  +  y  =  c  (1  -  xy).  15.  (x2  +  1)  (y^  +  1)  =  c. 

12.  V?/2  +  1  =  ex  -  y.  16.  {y  _  2)  (x  -  3)2  =  c  (x  -  4). 

13.  xy  =c  sin  x.  17.  xy  =  c  sin  y. 

EXERCISE  3 

1.  (6y +  2x)2(2/  +  x)  =  c.  4.  x2(x2  +  22/2)  =  c. 

2.  (7y  +  4x)2(2/  +  x)  =  c.  _j^- 

8.   lOgC  (5  y^  —xy  —  2x2)  5.    X  =  Ce     2a::2 

-     "^    iog^^^~'^~'^^^.     6.  x8  4- 3x2^  +  9x2/2 +  6y8_c, 
Vil        lOy  -  X  +  X  Vil       7.  x2 (x  +  3y)  =  c. 
43 


44  DIFFERENTIAL  EQUATIONS 


8.  Vx'  +  y2  —  x  +  c.  12.  x^  +  y^  =  cx^. 

9.  y  =  ce»*.  13.  y  =  cev^. 

10.  2x»  +  9x2|/  +  9x1/2  +  72/»  =  c.  14.  jS  _  ^/S  _  c^-j^. 

11.  x^  +  y2  =  ex. 

EXERCISE  4 

1.  (2x  +  6?/  +  8)2(x  +  2/  +  l)  =  c. 

2.  (X  +  72/  -  2)2(x  +  y  -  2)  =  c. 

8.  5x  + 152/ +  c  =  log(5x— 10J/  +  16). 
4.  2x  +  32/  +  c  =  log(3x-42/  +  2). 
6.  (x  +  2/  +  l)(5x+82/  +  14)2  =  c. 

6.  (X  +  2/  +  1)  (a:  +  52/  +  9)^  =  c. 

7.  logc(62/2  -  X2/  -  3x2  +  232/  -  8x  +  19 


9    ,     12  2/-x  +  23-(x  +  l)V73 
=  -p=log ^ -!—=• 

V73       122/-x  +  23  +  (x+l)V73 

8.  4x2— 10x2/ -3^2^  12x  + 22  2/  =  c. 

9.  7x2 -18x2/- 22/2 +  14x- 182/ =  c. 
10.  X  —  72/ +  c  =  log(3x  -  52/ +  11). 


EXERCISE  5 

1.  2x2  —  3x2/  +  42/2  =  c.  9.  x^y  —  Zxey ■{■  x^e^v  +  xy^  =  c. 

2.  7x2+ 16x2/ +  92/2  =  c.  10.  x tan 2/  =  c. 

3.  x2  + 3x2/ +  72/2  =  c.  11.  xsin-i2/  =  c. 

4.  x8  +  3x22/  =  c.  12.  2/  =  c (x  +  1)». 

6.  2x' +  9x22/ +  9x2/2  + 72/8  =  c.  13.  e2^  +  e^^  +  w  +  e2i' =  c. 

6.  x22/2  +  x2  +  2/2  =  c.  14.  2 x^2/^=j/e^  —  xe*'  +  c. 

7.  x*2/  —  x*2/2  =  c.  15.  Vx2  +  2/2  =  X  +  c. 

8.  C08X  •  C082/  =  c.  16.  x2  —  2/2  =  CX2/. 


EXERCISE  6 

1.  3x8  ^  Ty.2y-c,  6.  6xV  —  3x22/*  =  c. 

2.  x8  +  3x22/  =  c.  7.  x^j/^  +  xV^  =  c. 
8.  2xV  -  3x«2/«  =  c.  8.  a;2y2  ^.  ^  _  ^j.^ 

4.  x^2/^(2  X  -  3  2/)  =  c.  9.  x22/8  -x  =  cy^. 

6.  jc'l/(x  —  2/)  =  c.  10.  3x^2/7  —  5xV  =  c. 


ANSWERS  46 

EXERCISE  7 

1.  xy  =  x^  +  c.  7.  &'{xy  —  1)  =  c. 

2.  y  =  x{x^  +  c).  8.  2x^y  =  e^  +  ce-'^. 
8.  x^y  =  X  +  c.                                     9.  y  (a  +  1)  =  xg»  +  c. 

4.  y  =  sinx  4-  c  •  cosx.  10.  y{x  +  Vx^  +  1)  =  x  +  c. 

5.  2/(secx  +  tanx)  =  x  +  c.  11.  y  =  (x  +  c)e'»°-ix. 

6.  y  =  sinx  +  (x  +  c)cosx.  12.  4 y  Vx^  +  l  =  x^  —  2 x  +  cx-i. 


EXERCISE  8 

1.  y~^=zx{c  —  2x).  7.  x^  =  2/2(e^  +  c). 

2.  2/~^  =  x(x  +  c).  8.  x^  4-  2/3  =  ex. 

8.  x2|/3  =  x6  +  c.  9.  yi  -(x  +  c)  secx. 

4.  22/-3  =  x-H4e-+c)-(4e-+3x).  j^,   2y-*(2x+3)^  =  c(2x+3)-l. 

6.  X3  +  y3  ^  CX2.  ^  V       T     7 

EXERCISE  9 

1.  y  =  c  logxy.  2.  X  =  c  \ogxy. 

8.  (3x  +  2j/-5)^  =  c(2x-3y+  2). 

4.  3i/Vx-l  =  2(x-l)Vx2-l  +  cVx  +  l. 

5.  7/  =  x(eJ'  +  c).  18.  {by  +  o)«  =  c(ax  +  6)*. 

6.i/^  =  xMe^  +  c).  I9.sin?^  =  cx^. 

7.  y  =  x^{&=+  c).  X 

8.  eJ'  +  l  =  c(e^  +  l).  20.  y  =  cxcw. 

9.  e!'  =  e^  +  c.  21.  x^ys  =  e^  +  ce-'=. 

^  22.  sin^x  •  sin  y  =  c. 

10.  y  =  ce".  23.  X  sin?/  +  x^?/  —  y  cosx  =  c. 

11.  2x2  —  3x2/  +  62/2  =  c.  24.  x  +  y +  2-c{xy  +  x  +  y— 2). 


—  ne"^ 


25.  xlogxj/  =  cy. 

26.  x*  —  x^y'^  =  c. 


13.  sin 2/  =  c tanx.  •_i2'_     2 

14.  3  xSj/*  -  5  x22/6  =  c.  '*^-  ^™     X  ~  '^  • 

15.  log(x-22/+l)8+«  +  72/+c  =  0.    28.  x  =  2/2(x2  +  c). 

,„     .      ,2x  ,   ^  29.  X  — 3x2/2  + «^  =  c. 

16.  sm-i  —  —  x  +  c.  " 

,„     o       "^        o  30.  sin-ix2/  +  log^  =  c. 

17.  x2  —  X2/  =  C2/2.  °  X 


46  DIFFERENTIAL  EQUATIONS 

EXERCISE  10 

1.  {y  -  2x  +  c){y  -  4x  +  c)  =  0. 

2.  {y  -  4x  +  c){y  +  Sx  +  c)  =  0. 

3.  (2/  -  c)  (2/  -  4x  +  c)  (y  +  3x  +  c)  =  0. 

4.  {y-2x  +  c){y  +  x  +  c){2y-x  +  c)  =  0. 
6.  {y-2x  +  c)(y-3x  +  c){y-ix  +  c)  =  0. 

6.  (2/  —  ce-*)  (x2  +  2  J/  +  c)  =  0. 

7.  (x2  _  2 y  +  c)  (x2  -  1/  +  c)  (3x2  -  2 1/  +  c)  =  0. 

8.  y  —  x  +  c;  y  —  l  =  c{x  +  l). 

9.  (y  —  x  +  c)  (xy  —  ex  —  1)  =  0. 

10.  x2  +  (2/  -  c)2  =  1. 

11.  {xy  —  c)(y  —  cx)  =  0. 

12.  {xy  -  c)  (x2  +  y2  _  c)  =  0. 


EXERCISE  11 

2p       c  6.  (^x^  =  3cy  +  1. 

3       2)* '  7.  y  =  ex  +  c^ ;  singular  solution,  x^  +  4  y  =  0. 

p2      2  c  8.  1/  =  ex  +  tan-^c  ;  no  singular  solution. 

^'^J'^'J'  9.  2/  =  cx+6Vc2  +  l; 
2.  y  =  ce=^  +  c^.  singular  solution,  x^  +  y^  =  36. 

8.  1/  =  clog X  +  c".         10.  y  =  ex  +  - ;  singular  solution,  y^  =  4 ax. 

4.  y  =  ex^  +  e2.  c 

6.  ex*  =  (y  —  c)2.  11.  y  =  cx  +  cVcM^l. 

12.  y  =  ex  +  Ve^  +  1 ;  singular  solution,  x^  +  y^  =  1. 


EXERCISE  12 

1.  y^  =  cx  +  c2.  8.  y^  =  cx  +  cl 

2.  j/=p2_2p8  +  c;  4.  27CJ/2  =  8(x  — c)8. 
X  =  2i)  -  3p2.  6.  e2y  =  2cx  +  c^. 

a  y  =  c-p8-^-3p-31og(p-l); 

x  =  c-?|^-3p-31og(p-l). 

7.  X  =  cev  +  c2.  9.  e^  =  ce»  +  c^. 

8.  ci/8  =  (x  +  c)2.  10.  X  =  C2/2  +  c2. 


ANSWERS  47 

EXERCISE  13 

1.  x2  =  2cy  +  ci  12.  y^  =  cx  +  -^. 

2,  y  =  ex  +  logc.  c  +  1 

8.  X  =  ClOg2/  +  c2.  13.    (4x2  +  9y2_c)(x2  +  y2_c)^0. 

4.  J/  =  ex  ±  2  a  V^.  14.  (x  +  cf  +  y^  =  4.c^. 

5.  {x^  -  y^  +  c){y  -  ex)  =  0.  15.  (2x-3j/+c)(2x2-32/2+c)=0. 

6.  ?^  =  ex  ±  Va2c2  +  62.  16.  2/  =  ex  -  c2  +  c8. 

7.  eJ'  =  ce»  +  e2.^ 17.  (^y  _  c)(y  _  ce»)  =  o. 

i.  y  =  ex±  aVc2  +  l.  18.  (x?/^  -  e^x^y  —  c)  =  0. 

9.  2y^  =  S  cx2  +  c8.  19.  y^  +  c^  =  ex^. 

10.  2/2  =  2  CX2/2  +  c2x2.  20.  2/2e2»;  +  2  cy  +  c2  =  0. 

11.  (y-x2  +  e)(2/-logx  +  c)  =  0.    21    y2  =  cx2--^ 

1  +  c 

EXERCISE  14 

1.  y  =  Cje8^  +  626-^.  * 

2.  yz=  c^e*^  +  C2e2^.  6.  2/  =  Cj^e^''  +  c^^. 

3.  y  =  e^^^  +  egC-s^.  7.  y  =  0^62=^  +  Cgcs^  +  CgC*^. 

4.  y  =  CjC^^  +  Cgg-*^.  8.  2/  =  c^eS^  +  egC*  +  036-*==. 
6.  J/  =  Cj^e-**  +  Cg.  9.  y  =  e^e«*=  +  e^'^  +  036-"=. 

10.  2/  =  Cje*=*  +  C2e2^  +  Cgfi-  6^. 

EXERCISE  15 

1.  y  =  e3a^(Cj  cosx  +  CjSinx).  3.  y  =  e^  cos2x  +  e2sin2x. 

2.  y  =  e*^(c^  cos  2  X  +  C2  sin  2  x) .        4.  ?/  =  c^  cos  ox  +  ej  sin  ox. 

X     /  y —  I —       \ 

b.  y  =  Cje2^  +  e  ^1  Cg  cos x  +  Cg  sin x). 

6.  ?/ =  e  ^1  Cj  cos  —  x  +  egSin  —  x\- 

7.  y  =:  Cje2^  +  e-^(e2  cosVSx  +  CgSin  VSx). 

8.  J/  =  Cje2^  +  e2e-2a:  +  Cg  cos2x  +  C4Sin2x. 

X 

9.  y  =  e-^ff"^  +  Cge^  +  e*^(Cg  cosx  +  c^  sin  x). 

—  /  -\f%  "n/s    \       — /  "\/^  "x/^   \ 

10.  y  =  e'*(cjCos  —  x  +  e2sin  —  xj+ e  ^(cgcos  —  x  +  c^sin  —  xj. 


48  DIFFERENTIAL  EQUATIONS 

EXERCISE  16 
1.  y  =  e-2x(CjX  +  C2).  J 

S.y  =  e2^(CiX  +  C2)  +  Cg^.  »•  J'  =  ^-8=^(013;  +  Cj)  +  CjC^. 

4.  2/  =  e2^(CjX  +  C2)  +  CgX  +  c,.  9'  y  =  ^(<^i^  +  C2)  +  ^3^-8'. 

6.  y  =  Cie8-+C2x8  +  C3x2  +  c,x  +  C5.  10-  1/ =  c^ ^ (c^x^  +  C^X  +  C3) . 
6.  y  =  e8^  (c^x*  +  CgX  +  Cg)  +  c^e*  =^. 

EXERCISE  17 

1.  y  =  ^{CjX^  +  CgX  +  Cg)  +  e2x. 

2.  y  =  Cie^"  +  Cj  cos  VSx  +  Cg  sin  VSx  —  e^'^. 

3.  y  =  c^esx  +  c^ea^  +  CgC^^  +  J  e*^.  ^.^^ 

4.  2/  =  Cie2^  +  C2e-2a:  +  Cg  COS  2  X  +  c^sin2x . 

5.  |/  =  Cie8=f  +  C2e9^  +  3x-5.  ^^ 

6.  y  =  CjC**  +  e^(C2  — "^x). 

T.  y  =  0^0^+  c^e-'^  +  2x^  —  ^x  —  4. 

8.  y  =  Cj  cos  2 x  +  C2  sin  2 X  +  \x  cos 2 x. 

o        W        »\  .  cosSx— 7sin3x      e^^^,,^  .    „       „       „  ^ 

9.  i/=Cje-3^+e<^(c2+-)  + -— — — -(19sin3x  +  9co83x), 

\         7/  160  442 


2x 


10.  y  =  Cie8  +C2e2  +  ^. 


EXERCISE  18 


1.  xy  =  Cj  logx  +  Cj.  3.  1/  =  CjX  +  C2X'*  +  CgX'. 

x^  c 

2.  y  =  CiX  +  C2X*-— •  4.  2/  =  Cix2  +  C2X*  + J. 

5.  y  =  -i  +  x(c2  cos  Vs  logx  +  Cg  sin  Vs  logx)  +  i  logx. 

x^ 

6.  x<y  =  Cjxs  +  CgX^  +  CgX  +  c^.       8.  r/  =  c^{ax  +  6)*  +  C2(ax  +  6)-«. 

7.  8  =  Ci«  +  C2-fclog«.  9.  y  =  Ci(2x-3)  +  C2(2x-8)-*. 

10.  y  =  Cj(x  —  1)2  +  Cg  cos  log(x  —  1)^  +  Cg  sin  log(x  —  1)^. 

EXERCISE  19 

1.  y  —  c^  logx  +  Cj.  6.  y  =  X  +  C2  —  cos(x  +  c^). 

2.  y  =  Cj  sin-ix  +  Cj.  7.  y  ='Cj  sec-^x  +  C2. 

3.  1/'=  logsec(x  +  Cj)  +  Cj.  8.  y  =  c^  secx  +  c,. 

4.  2/  =  log(x  +  Ci)  +  Cj.  9.  y  =  (X  +  Cj) log(x  +  Cj)  +  Cj. 
6.  J/  =  c^  tanx  +  Cj.  10.  (x  —  Cj)2  +■  {y  —  c„)^  =  a*. 


ANSWERS  49 

EXERCISE  20 

1.  (X  +  Ci)2  +  2/2  =  d  6.  y  =  Cj  sec  (X  +  c^). 

2.  e!'  =  Cj  sin x  +  Cg  cosx.  7.  yi  =  Cj  sin  (2x4-  c^)- 
S.  y  =  c^tP^.  8.  sin  y  =  c^e"^  +  c^e-''. 
4.  log?/  =  Cie2^  +  036-2 a;.  9.  tany  =  CjC^^  +  Cge-^. 

10.  a sin-iy  =  x  +  Cji/  +  Cg. 


a"^ 


EXERCISE  21 


2.  ?/  =  CiSin(ax  +  C2).  4.  x  +  c^zz  —  log     , -■ 

8.  (X  +  C2)2  -  c2j/2  =  c*.   ''''^       Veiz  +  cf  +  ci 

6.  2(2/?  -  2ci^)  \A/^  +  cf  =  3x  +  c^. 

6.  c^v^  Vj/  _  2 cf  +  2  cf  log (2/^  +  V2/  -  2  cf)  =  X  +  Cg. 

7.  4  0^026  2  =  'icle''-^  +  ae~*^2^. 

8.  2/*\/32/t  +  Scf  -  cf  V3  logU^  +  Vyt  +  cl)=  2x  +  03. 

9.  2(3 yt  _  4c,2/^  +  Scf)  Vyi  +  Cj  =  5V6x  +  c^. 
10.  sini/ =  V 1  + cf  sin ?• 


EXERCISE  22 

1.  (x2  -  3x)2/  =  e^  +  x3  +  Cjx2  +  c^x  +  Cg. 

2.  (2x  +  5)2y  =— sinx  +  Cj^x  +  Cg.     • 

X 

5.  y  sin  X  =  4  e2  +  c^x  +  Cj . 

4.  2/Vl  +  x2  =  (1  +  x2)f  +  Ci log(x  +  Vl  +  x2)+  Ca 

**-^  =  3^  +  '- 

6.  (3x2  — 6x  + 7)i/  =  e^  +  CiX  + Cg. 

7.  x'y  —  Cjx2  +  CgX  +  C3  +  3  e*  +  cosx. 
%.  y  —  cosx  +  e2^  +  Cix2  +  CgX  +  Cg. 
9.  e=^(2/ +  C]X  +  Cg)  =  X  +  Cg. 

X 

10.  y  =  a  sin-i  — I-  CjX  +  Cg. 


50  DIFFERENTIAL  EQUATIONS 

EXERCISE  23 

1.  y  =  CjSinx- sin(2x  +  a).  6.  j/sinox  =  CjC*"  +  CjC-**". 

2.  y  =  e^{c^cos3x  +  c^ehiSx).      ^'  y  =  x^{c'co82x  + c,sm2x). 
S.y  =  sin  X  (c^e-  +  c^)-  9.  y  =  log x  •  (c^^  +  c^e- -) . 

4.  y  =  e=^(CjX*  +  CgX^) .  2Q   j^  log  X  =  Cj  cos  ax  +  c^  sin  ax. 

5.  y  =  X2  (Cj  cos  3  X  +  Cg  sin  3  x). 

EXERCISE  24 

1.  y  =  Cj sin [2 log (secx  +  tanx)  +  a].  6.  y  Vx^  +  a^  —  CjX  +  c^. 

2.  i/  =  c,xVl-x2  +  C2 (1-2x2).  7.  y  =  CiSinx  +  C2Cscx. 

ax^  cue* 

5.  1/ Vx2  +  I  =  CjX  +  Cg.  S.  y  =  c^e^  +  Cge'^s". 
4.  2/  =  Cj  sec  X  4-  Cg  cos  x.                              g_  ^  _  ^^  ggg  x  +  c,  tan  x. 

6.  xj/  =  Cj  +  C2 Vx2  —  1.  10.  2/  =  CjC^  +  CgC-^^. 

EXERCISE  25 

1.  1/  =  c^ef^  +  CgC-*".  9.  2/  =  Ci (x2  +  a2)»  +  Cg. 

2.  y  =  ±  cos(x  +  Cj)  +  Cg.  10.  j/ =  e-2a;(CjCos2x  +  CgSin  2x). 
8.  i/  =  CiX  +  C2  +  e^='(c3X  +  cJ.       11.  j/  =  Ci(2x-3)''+C2(2x-3)-''. 

4.  xy  +  CjX  +  c^y  +  03  =  0.  12.  j/  =  e^ (c^  sin x  +  c,). 

5.  y  =  CjX  +  C2  Vl  —  x2.  13.  y  =  0^+  c^^'^  +  Cge-  8^. 

6.  y  =  Ci  sec*x  +  Cg  cosx.  14.  j/  =  Cj  tan  x  +  Cg. 

7.  sin  J/  =  CjX  +  C2.  15.  2/  =  Cj  sin^x  +  c^  sin  x. 

8.  y^  +  CjX  +  c^y  +  Cg  =  0.  16.  t^  =  c^eJ'  +  CgC-w. 

17.  y  =  c^  (loga;)2  +  C2  (logx)^. 

18.  y  =  e='(qcosx  +  CgSinx)^-  e-^(c3C0sx  +  c^sinx). 

19.  2/=e6a:(Ci+x)  +  efi^(c2  +  x).     21.  tan-^y  =  c.^ef'+ c^e-^^. 

20.  et>  =  c^e^  +  026-^.  22.  s  =  c^  sin  (of  +  «)• 

23.  y  =  e2a:(CiC0s3x  +  C2  8in3x)+  e8a:(c3COs2x  +  c^sin2x). 

24.  (logi/)2  =  Cje2«=+  CjC-^o^.       27.  1/  =  e<"(CjCosax  +  CgSinox). 
26.  x6j/  =  Cj  +  CgX  +  CgX".  28.  e^  =  CiSinx  +  c^. 

26.  J/  =  Cj  +  e2^(c2X  +  Cg).  29.  y-  Vx^  +  a^  +  CjX  +  Cj. 

30.  2/  =  c,e^+  e-8=»^(c2X  +  Cg)+  6-8^(046-^^+ Cgg--^^). 


ANSWERS  51 

EXERCISE  26 

l.y^  =  4:kx  +  c.  ^.y  =  kx^+c.  7.  4x2+ 9^/2  =  c^. 

2.  y^  =  2 kx^  +  c.  5.  y^  =  2kx  +  c.  6.  x^+y^  =  c^. 

8,  y'^  =  x^+  c2.  6.  y  =  ce*=^. 

9.  (x  -  yf-2k{x  +  y)  +  k'^  =  0. 

10.  xy  =  k^.  11.  xy  =  c^.  12.  y  =  ex. 

13.  c^s^  =  k  +  (c^t  +  Cg)^. 

14.  Cj  VcjS^—  2 A;s  +  2 A;  Vc^ log( Vqi  +  Vc^s - 2 jfc)  =  Ci^i  +  c^. 

15.  s  =  c  sin  (fc^  +  «)  • 

16.  (X  -  Ci)2  +{y-  C2)2  =  a2. 

17.  {x  +  c,)^  =  c^{2y-c^). 

18.  y  +  c.^  =  u  +  klog^r^-^  .   [m2  =  ^2 -  (x  +  c^)^.] 

\  u  +  k 

19.  Ely  ::.^C^ -%)-'""'' 


2  \  8        12/        384 
32  ■ 


„„    „^        P/te2      x8\ 

20.  EIy  =  -{ ) 

2\4        6/ 

-,     _,^        w  ,,        ...  wl^x      wl* 

21.  JSTw  =  —  (i  —  x)*  + 

24^  '6  24 

22.  ^72/  =  _(Z_x)3+-^ -. 

V         -^t 

23.  C  =  -  +  ce   -f'   . 

«  _^ 

_  J_      p    RK  r>  ~ 

24.  C  =  ce   ^*+  ^ /  eiiKf'{t)dt. 

t 
'        p~TiK  r  — 

25.  q^ce   '«^+ /  eR«f{t)di. 


ANNOUNCEMENTS 


BOOKS   IN 
HIGHER   MATHEMATICS 

ADVANCED  CALCULUS 

By  Edwin  Bidwell  Wilson,  Massachusetts  Institute  of  Technology. 
A  COMPREHENSIVE  second  course  in  calculus.    Throughout,  attention 
has  been  paid  to  the  needs  of  students  of  engineering  and  mathematical 
physics.    There  is  an  abundance  of  problems.  ^66 pages,  t>s.oo. 

SYNTHETIC  PROJECTIVE  GEOMETRY 

By  Derrick  Norman  Lehmer,  The  University  of  California. 
A   SIMPLE  treatment  which  avoids   algebraic  methods,  in  favor  of 
purely  synthetic.    The  book  emphasizes  especially  the  unity  and  sym- 
metry underlying  the  subject.    i2j  pages,  96  cents. 

ANALYTIC  GEOMETRY  AND  CALCULUS 

By  Frederick  S.  Woods  and  Frederick  H.  Bailey,  Massachusetts  Institute  of 
Technology. 

A  TWO-YEAR  college  course,  notable  because  at  an  early  point  in  the 
treatment  it  intermingles  the  methods  of  analytic  geometry  and  the  calcu- 
lus.   The  course  provides  over  two  thousand  problems.  ^16 pages,  $j.oo. 

THEORY  OF  MAXIMA  AND  MINIMA 

By  Harris  Hancock,  University  of  Cincinnati. 
It  aims  to  provide  for  American  students  an  adequate  account  of 
the  theory,  freed  from   current  errors.     The   book  contains  a  great 
number   of   problems    and   exercises   taken   from   algebra,   geometry, 
mechanics,  and  physics.     [In  preparation] 

ELEMENTARY  COURSE  IN  DIFFERENTIAL  EQUATIONS 

By  Edward  J.  Maurus,  Notre  Dame  University. 
An  easy  introduction  into  differential  equations,  primarily  for  the 
use  of  students  in  engineering  courses.    No  attempt  is  made  to  adhere 
to  rigidity.    Numerous  problems  are  included. 

INTRODUCTION  TO  THE  ELEMENTARY  FUNCTIONS 

By  Raymond   Benedict  McClenon,  Grinnell   College.      Edited  by  William 
James  Rusk,  Grinnell  College. 

A  UNIFIED  course  for  freshman  classes  in  colleges  and  technical 
schools.  It  includes  the  most  important  topics  usually  treated  under 
trigonometry  and  elementary  analytic  geometry,  with  a  simple  introduc- 
tion to  differential  calculus. 


'33  ^ 

GINN  AND   COMPANY  Publishers 


RECENT   BOOKS   IN   HIGHER 
MATHEMATICS 


COURSE  IN  MATHEMATICAL  ANALYSIS 

By  Edouard  Goursat.  Translated  by  E.  R.  Hedrick  and  Otto  Dunkel,  Uni- 
versity of  Missouri.  Volume  1,  J4.00;  Volume  II,  Part  I,  $2.75;  Volume  II 
Part  II,  #2.75. 

A  STANDARD  French  work  well  known  to  American  mathematicians  for  its 
clear  style,  its  wealth  of  material,  and  its  thoroughness  and  rigor.  Volume  I 
covers  the  subjects  of  a  second  course  in  calculus.  Parts  I  and  1 1  of  Volume  1 1 
treat  respectively  functions  of  a  complex  variable  and  differential  equations. 

PROJECTIVE  GEOMETRY 

By  Oswald  Veblen,  Princeton  University,  and  J.  W.  Young,  Dartmouth  College. 
Volume  I,  f4.oo;  Volume  II  [/n press]. 

A  THOROUGH  exposition  that  takes  into  account  modem  concepts  and  methods. 
Volume  II  deals  with  important  special  branches  of  geometry  obtained  by 
adding  further  assumptions  or  by  considering  particular  subgroups  of  the 
projective  group. 

THE  THEORY  OF  INVARIANTS 

By  Oliver  E.  Glenn,  University  of  Pennsylvania.   $2.75. 
Ten  chapters  giving  an  adequate  introduction  to  the  study  of  original  memoirs. 
Both  the  symbolical  and  the  nonsymbolical  methods  are  employed. 


THE    USE    OF     GENERALIZED     COORDINATES    IN 
MECHANICS  AND  PHYSICS 

By  William  E.  Byerlv,  Harvard  University.   $1.25, 
A  BRIEF,  practical  introduction  with  numerous  illustrative   examples.   The 
Lagrangian  and  Hamiltonian  equations  of  motion  are  obtained  and  applied. 

PROBLEMS  IN  THE  CALCULUS 

By  David  D.  Lkib,  Yale  University.   J  1.00. 
Over  two  thousand  carefully  organized  problems,  covering  the  entire  field 
of  calculus. 

SOLID  GEOMETRY 

By  Sophia  Foster  Richardson,  Vassar  College.  90  cents. 
A  college  course  with  emphasis  on  original  work.   The  methods  of  conti- 
nental Europe  receive  consideration. 


GINN  AND   COMPANY  Publishers 


University  of  California 

SOUTHERN  REGIONAL  LIBRARY  FACILITY 

Return  this  material  to  the  library 

from  which  it  was  borrowed. 


_!£21!iP*'"*  REGIONAL  UBRAfiY  FACILITY 


A     000  169  533     7 
Engineering  & 
Mathematical 

Sciences 

Ubraqr 


311 


Auxrawn 


/2 


